- Final Track Plinth Design -

Document Reference: HM-LE-PRWIDE-M0601-DOC-DC-TRW-00-101-C

Date Revision subject

Issue no. 1 10/10/2012 First Issue
Issue no. 2 11/01/2013 Second Issue – CE’s Comment Incorporation
Issue no. 3 04/03/2013 CE’s Comment Incorporation
Issue no. 4 15/05/2013 Clients comment Incorporated
Change in track plinth geometry consequent to
Issue no. 5 11/07/2013
HMR comments
Issue no.6 21/08/2013 PC’s Comment Incorporation
Incorporate CE’s comment as per Doc No. HM-LE-
Issue no.7 18/09/2013
PRWIDE-M0601-DOC-DC-TRW-00-101-A
Issue no.8 12/11/2013 I.E.’s comment Incorporate

Issue no.7 Written by Signature

P Mandley

Checked by Signature

S. Biswas

Validated by Signature

Jean Pierre BERGOEND

Chapter Page Modification

7 13 Load Factor changed
8.1 15 Fy changed in SLS

SYSTRA MVA Consulting (India) Private Limited Telephone: 91 11 26422844

5th Floor, Guru AngadBhawan, 71, Nehru Place, New Delhi 110019 Fax: 91 11 26224204

The copyright of this document belongs to SYSTRA

Table of contents

1 PURPOSE 4
2 ABBREVIATIONS 4
3 SCOPE 5
4 REFERENCE AND STANDARDS 6
4.1 REFERENCE DOCUMENTS 6
4.2 STANDARDS AND NORMS 6
4.2.1 Indian Railway Standards 6
4.2.2 Indian Standards (IS) 6
4.2.3 Other Codes and Standards 6
5 DESIGN CRITERIA 7
5.1 CODES AND STANDARDS 7
5.2 MATERIAL CHARACTERISTICS 7
5.2.1 Concrete 7
5.2.2 Steel reinforcement 7
5.3 TRACK PLINTH DEFINITION 7
6 LOADS 8
6.1 LOADS LIST 8
6.1.1 Train Load 8
6.1.2 Impact load (DY) 8
6.1.3 Self-weight of Rail & Fastening (SIDL) 8
6.1.4 Self-weight of rail plinth 8
6.1.5 Lurching Force 8
6.1.6 Racking force 8
6.1.7 Centrifugal Force 9
6.1.8 Wind Load 9
6.1.9 Braking and Traction Load 9
6.1.10 Derailment Load 10
6.1.11 Thermal expansion of a curved rail 10
6.1.12 Long welded rail forces (LR) 11
6.1.13 Earthquake (EQ) 11
6.2 LOADS COMBINATIONS 12
7 FORCES SUMMARY 13
7.1 MAXIMUM FY CASE 13
7.2 MINIMUM FY CASE 14
8 SHEAR CONNECTOR DESIGN 15
8.1 DESIGN LOADINGS 15
8.2 CONCRETE SECTIONS STUDIED 15
8.3 REINFORCED CONCRETE SECTIONS CHECK 16
8.3.1 First step: Shear Transfer Check 16
8.3.2 Second step: Section under Flexure – Compression Check. 17
8.4 SHEAR TRANSFER CHECK (FOR FORMULA AND REFERENCES REFER CL NO. 8.3.1) 18
8.5 SECTION UNDER FLEXURE-COMPRESSION CHECK 19
8.5.1 ULS combination: 19
8.5.2 SLS combination: 21
9 DESIGN OF DERAILMENT GUARD 23
9.1 REINFORCEMENT CALCULATION FOR DERAILMENT GUARD 24
9.2 SHEAR REINFORCEMENT FOR DERAILMENT GUARD 25
9.3 REINFORCEMENT CONCLUSION 25
9.3.1 Flexure reinforcement 25
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 9.3.2 Shear reinforcement 25
10 PLINTH REINFORCEMENT DESIGN 26
10.1 LONGITUDINAL STEEL BARS 26
10.2 TRANSVERSAL FRAMES 26
10.3 SHRINKAGE & TEMPERATURE CHECK 27
11 TRACK PLINTH WITH CHECK RAIL 29
11.1 SHEAR CHECK SLS CASE 30
11.2 SHEAR CHECK ULS CASE 31
11.3 SECTION UNDER FLEXURE-COMPRESSION CHECK 32
11.3.1 ULS combination: 32
11.3.2 SLS combination: 34
11.4 REINFORCEMENT CALCULATION FOR DERAILMENT GUARD 36
12 PLINTH REINFORCEMENT DESIGN 37
12.1 LONGITUDINAL STEEL BARS 37
12.2 TRANSVERSAL FRAMES 37

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1 PURPOSE

The purpose of this Design Calculation Note is to provide detailed design of concrete track plinth supported over
box girder and also check the plinth shear connectors for Hyderabad Metro Rail Project.

2 ABBREVIATIONS
ASTM American Society for Testing and Materials
BS British Standards
CW Civil Works
CWR Continuous Welded Rails
EN European Norms
ER Employer’s Representative
ERQ Employer’s Requirements
IRS Indian Railway Standards
ISO International Standards Organization
kN Kilo Newton (unit)
LWR Long Welded Rail
m meter (unit)
mm millimeter (unit)
R Radius
RSI Rail - Structure Interaction
SoD Schedule of Dimensions
TMT Thermo Mechanically Treated
TOR Top Of Rail
UIC International Union of Railways

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3 SCOPE
The Hyderabad Metro rail Project has been planned with total 3 corridors having a total number of 65 stations and
track of 71.6 Route Kilometers (divided into 6 stages).
The entire route will be an elevated corridor. The scope of work includes the Detail Design Consultancy for the
Track works in Viaduct as well as Depot.

Corridor Name Route km Stations Depot

Corridor-I Miyapur- L.B. Nagar 29.21kms 27 elevated Miyapur (Depot-I)

stations

Corridor-II Jubilee Bus Station – 15.18 kms 15 elevated Falaknuma(Depot-II)

Falaknuma stations

Corridor-III Nagole – 27.29 kms 23 elevated Uppal(Depot-III)

Shilparamam stations

The present Document is prepared under the scope of Trackwork Detailed design consultancy for the Hyderabad
Metro Rail Project. This Design calculation is applicable to track plinth on viaduct on straight as well as on curve
alignment and deals with:
Structural design of standard track plinth& track plinth with check rail,
Structural design of derailment guard,
Check for the shear connector between track plinth and viaduct.

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4 REFERENCE AND STANDARDS
4.1 Reference documents
The following documents are the reference for design.

R1 Technical Specifications Trackwork

R2 Manual of Standards and Specifications
R3 DBR – Section 6 – Trackworks
R4 DBR – Section 2 – Viaduct

4.2 Standards and Norms

Whenever they exist, regulations and standards related to layout must be used, preferably in this order:
- European Norms (CEN),
- Indian Railway Standards (IRS)
- UIC standards,
- ISO standards,
- National standards.

4.2.1 Indian Railway Standards

IRS Concrete Bridge Code: 1997.
Bridge Rules - Rules specifying the loads for design of super-structure and sub-structure of bridges and for
assessment of the strength of existing bridges (second reprinting: 2008).
Indian Railway Manual of instructions on Long welded rails 1996.

4.2.2 Indian Standards (IS)

IS-432 Part 1-1982 Mild steel and medium tensile steel bars.
IS-456:2000 Code of practice for plain and reinforced concrete.
IS-875 Code of practice for design loads.
IS-1786:2008 High strength deformed steel bars and wires for concrete reinforcement - Specification.
IS-2502:1963 Code of Practice for Bending and Fixing of Bars for Concrete Reinforcement.
Report No. BS-14 of March, 1999 - Durability of Concrete Structures.

4.2.3 Other Codes and Standards

IRC-6:2010 Standard specifications and codes of practice for road bridges
AASHTO:1996 Guide specification for design and construction of segmental concrete bridges.
ACI 358.1R-92 Analysis and design of reinforced concrete - Concrete guide way structure.
BS-5400, Part2:1978 Steel, concrete and composite bridges, Part 2 : Specification for loads.
EN1991, Part2: 2003 Eurocode 1, Actions on structures, Traffic loads on bridges.
BD 28-87: Early thermal cracking of concrete

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5 DESIGN CRITERIA
5.1 Codes and standards
Basis characteristics and formulas are given as per following codes and standards.
IRS Concrete Bridge Code: 1997.
AASHTO: 1996 Guide specification for design and construction of segmental concrete bridges.

5.2 Material characteristics

5.2.1 Concrete
Characteristic concrete strength is taken as:
fck = 35 MPa
3
Density of reinforced concrete is taken as 24 KN/m .
Allowable compressive stress for concrete in SLS is 0.50 fck (Refer IRS:CBC cl.no. 10.2.2.1)

Concrete cover
Concrete cover is 40 mm for plinth.
5.2.2 Steel reinforcement
Yield stress is taken as:fy = 500 MPa.
Young modulus is taken as: Ec = 200,000 MPa.
Allowable tensile stress for reinforcement in SLS is 0.75 fy (Refer IRS:CBC cl.no. 10.2.2.1)
Modulus of elasticity of rail Erail =210,000 MPa (Refer RSI analysis)

5.3 Track Plinth Definition

The plinth dimensions vary all along the track. In curve alignment, the minimum dimension shall be followed for
the innermost track plinth and the dimension for outer track plinth shall be calculated as per relative super
elevation.
The Geometry of the track plinth is variable:

Min. Thickness of base slab of track plinth = 0.272 m

Min. Width of base slab of track plinth = 0.730 m
Max. Length of track plinth segment = 4.350 m
Spacing of Fastener = 0.650 m
The typical cross section of track plinth is as shown in following figure:

Typical track plinth cross-section

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6 LOADS
The forces of the train wheel on the rail are required to design the track plinth reinforcements. Therefore all the
following loads are calculated at the Wheel / Rail contact on one single rail.
The axis (X, Y, Z) are shown on the following drawing.

Axis convention at the rail top

6.1 Loads list
6.1.1 Train Load
As per DBR cl. no. 5.6.1, the vertical load due to live load is 170 kN/axle.
It is assumed that the single wheel load is directly transferred to the effective cross section area as mention in cl.
no. 5.3.
Hence Vertical force per wheel is Fy = 170 kN/2 = 85 kN. (For one track plinth)

6.1.2 Impact load (DY)

6.1.3 Self-weight of Rail & Fastening (SIDL)

As per DBR cl. no. 5.5.1, Self-weight of Rail and fastening system is = 3 kN/m.
Hence SIDL per fastening is Fy = 3*0.65/4 = 0.49 kN/fastening.

6.1.4 Self-weight of rail plinth

As per DBR cl. no. 5.5.1,self-weight of the track plinth is 28 KN/m
Self-weight of the track plinth/fastening is:Fy = 28/4*0.65 = 4.55 kN/Fastening.

6.1.5 Lurching Force

Lurching forces are caused by the train rotating slightly about its axis.
As per BS 5400-Part 2, clause No, 8.2.7, 56% and 44% of the train load shall be applied. This causes a moment
at rail level corresponding to 6% of the maximum axle load multiplied by the distance between rails. Hence,
Vertical force variation: Fy = +/- 6 % * 170 = +/- 10.2 kN/Wheel.
6.1.6 Racking force
The nosing force (racking force) shall be taken as a force acting horizontally, at the top of the rails, perpendicular
to the center-line of track. It shall be applied on both straight track and curved track. The nosing force shall always
be combined with a vertical traffic load.
As per Bridge Rules, clause No. 2.9.1, the racking force value to be taken into account, is 5.88 kN/m
Hence racking force per fastening system is: Fx= 5.88 * 0.65 = 3.83 kN/fastening.
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6.1.7 Centrifugal Force
The centrifugal force is different for the different radius of curvature of the alignment. The comparative analysis is
done for the worst case consideration as follows:

The rail on curve alignment of radius 250m with maximum speed of 70 km/hr. produces maximum effect of
centrifugal force. Lever arm for centrifugal force as per bridge rule cl no. 2.5.3(b) is 1.83m above rail level.
2 2
Horizontal force: LL*V 170 * 70
Fx = = = 26.24kN.
127*R 127 * 250

Vertical force: Fy= ( 26.24 x 1.83) / 1.435 = +/- 33.46kN.

6.1.8 Wind Load

For structure with live load, as per DBR cl no. 5.17, wind load as per IRC6:2010 cl. no. 209 is,
FT = Pz x A1 x G x CD
Train height above rail level = 4.048m
Parapet height above rail level = 1.095m

Where,
2 2
Pz = Hour mean wind speed in N/mm , = 230.5 N/mm (Terrain with obstruction, 15m above GL)
2 2
A1 = Solid area in m = (2.15*2 + 2.2*2 + 12.4) * (4.048-1.095) = 62.3m
G = Gust factor = 2
CD = Drag coefficient = 1.5 (Max)
Total horizontal force = 230.5*62.3*2*1.5 = 43081 N = 43.1 kN
Horizontal Force/ axle Fx = 43.1/4 = 10.8 kN/axle
Lever arm for wind load above rail level = [(4.048-1.095)/2+1.095] = 2.57m

Moment induced by wind load at the rail top level: Mx = 10.8 * 2.57 = 27.8 kN.m.
So we can calculate the vertical force on one rail to balance this moment: Fy = 27.8 / 1.435 = +/- 19.38 kN.

6.1.9 Braking and Traction Load

The maximum of braking or traction force is considered for the design of one track plinth.
It is assumed that longitudinal force due to braking and traction is uniformly distributed along length of wagon.
The length of one wagon is 21.1m
No. of axle in one wagon is 4 nos.
Longitudinal Force: 20% x Total wheel load per wagon per rail / Length of wagon (As per DBR cl no. 5.9.2)
Fz = 20% *85*4/21.1= 3.22 kN/m
Or
Fz = 3.22 * 0.65 = 2.1 kN/fastening

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6.1.10 Derailment Load
For derailment check, we intend to use ACI 358.1R92, clause No. 3.5.2 which corresponds to the application of
50% of one vehicle weight (live load) applied horizontally as a 5m long uniform impact load on member used as
derailment wall.
Hence, per 5 meter length: Fx= 170*4*0.5 = 340kN/ 5m.
Hence: Fx = 340/5 = 68 kN/m.
Derailment load on effective width per fastening is: 68 * 0.65 = 44.2 kN/fastening.

6.1.11 Thermal expansion of a curved rail

A curved rail subjected to temperature change will induce a radial force on the plinth, which depends on the rail
curvature.

Stress
in rail

Radial force
from plinth
Stress
in rail

Balance of a curved rail through thermal expansion

Horizontal force due to thermal expansion is calculated same as centrifugal force (assuming the temperature
variation of +/- 30°C as per DBR cl. no. 5.12.3):

7680 x 210000 x 0.012 E-3 x 30

= 4.85 kN/m.
120

For a 4.35-meter long plinth segment (this force is transmitted by the rail fasteners to plinth):
Fx = 0.65 * 4.85 = 3.2 kN/Fastening.

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6.1.12 Long welded rail forces (LR)
The maximum longitudinal force induced on plinth by the LWR force is limited to the longitudinal restraint capacity
of the fastening system which is: LR = 15.1kN/fastening (As per repeated load test results provided by L&T).

Fz = 15.1 kN/fastening.

6.1.13 Earthquake (EQ)

As per DBR cl no 5.18.8, the earthquake load in transverse direction due to live load is:
Ah = (Z/2) * (I/R) * (Sa/g), where:
Ah = horizontal seismic coefficient to be considered in design (Seismic acceleration = Ah*g)
Z = Zone factor = 0.10 (Hyderabad region = zone II)
I = Importance factor = 1.5
R = Response Reduction factor = 4.0 (for RCC substructure)
Sa/g = 2.5 (Considering max. value)
Ah = (0.1/2) * (1.5/4) * (2.5) = 0.047
Fx = 0.047*170*25/100 = 2kN/wheel.
(As per viaduct DBR cl no. 5.18.6, 25% of train mass is considered while evaluating seismic force in transverse
direction)

The vertical component due to earthquake force is

Fy = (Fx) x (Lever arm for EQ force) / (c/c dis. between two rail)
Fy = 2*1.83/1.435 = 2.55kN/wheel.
Note: Lever arm for EQ force is assumed same as centrifugal force.

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6.2 Loads combinations
The following load combinations are proposed as per DBR cl. no. 6.

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7 FORCES SUMMARY
7.1 Maximum FY case

Limit Load factor

Loads
state G-I GII-A GII-B G-III G-V
ULS 1.25 1.25 1.25 1.25 1.25
Dead load
SLS 1 1 1 1 1
ULS 1.75 0 0.7 0 0
Live load
SLS 1.1 0 0.5 1 0
Lurching ULS 1.75 0 0.7 0 0
Force &
Rack ing force SLS 1.1 0 0.5 1 0

Centrifugal ULS 1.75 0 0.7 0 0

force SLS 1.1 0 0.5 1 0
Breaking & ULS 1.75 0 0.7 0 0
Traction SLS 1.1 0 0.5 1 0
ULS 2 2 2 0 2
SIDL
SLS 1.2 1.2 1.2 1.2 1.2
Radial force ULS 0 0 0 0 0
due to temp. SLS 0 0 0 1 0
ULS 0 1.6 1.25 0 0
Earthquake
SLS 0 1 1 0 0
ULS 0 0 0 0 0
LWR
SLS 0 0 0 1 0
Derailment ULS 0 0 0 0 1.75
load SLS 0 0 0 0 1
ULS
Wind Load
SLS

While calculating the forces for the analysis of track plinth, it is assumed that the all-wheel/axle loads are directly
transferred to the single fastener. As the live load and wind load acting over live load is transferred to the track
plinth through wheel/axle, hence these loads are calculated per wheel or axle load. Few forces like dead load,
SIDL, LWR or derailment load etc, are uniformly distributed over length of rail/plinth and hence these forces are
calculated per fastener. Hence forces per wheel/axle are directly added to forces per fastener.

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7.2 Minimum FY case

Limit Load factor

Loads
state G-I GII-A GII-B G-III G-V
ULS 1.25 1.25 1.25 1.25 1.25
Dead load
SLS 1 1 1 1 1
ULS 1.75 0 0.7 0 0
Live load
SLS 1.1 0 0.5 1 0
Lurching ULS 1.75 0 0.7 0 0
Force &
Rack ing force SLS 1.1 0 0.5 1 0
Centrifugal ULS 1.75 0 0.7 0 0
force SLS 1.1 0 0.5 1 0
Break ing & ULS 1.75 0 0.7 0 0
Traction SLS 1.1 0 0.5 1 0
ULS 2 2 2 0 2
SIDL
SLS 1.2 1.2 1.2 1.2 1.2
Radial force ULS 0 0 0 0 0
due to temp. SLS 0 0 0 1 0
ULS 0 1.6 1.25 0 0
Earthquak e
SLS 0 1 1 0 0
ULS 0 0 0 0 0
LWR
SLS 0 0 0 1 0
Derailment ULS 0 0 0 0 1.75
load SLS 0 0 0 0 1
ULS
Wind Load
SLS

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8 SHEAR CONNECTOR DESIGN
The design is carried out at the plinth bottom on a reinforced concrete section resulting from the rail force spread
from rail to plinth through fasteners.

8.1 Design loadings

According to the table on paragraph § 7.1 &7.2, the factored forces and moment for design at the plinth bottom
are as follows:
It is assumed that the horizontal forces are acting at rail top. Total height of rail top from the bottom of track plinth
is (290+211) =501 mm as shown in cl .no. 5.3.
Maximum cant for 120m radius is = 125 mm
Hence lever arm for horizontal forces = 501 +125 = 626 mm ≈ 0.630m

ULS combination :
Fx = 77.40kN/Fastening
Fy = 301.7kN (Max)
Fy = 5.700kN (Min)
Fz = 3.700kN
Mz = 48.80kN-m (77.4 *0.630) (Force * Lever arm)
SLS combination :
Fx = 44.20 kN/m
Fy = 190.6kN (Max)
Fy = 5.140kN (Min)
Fz = 17.20kN
Mz = 27.85kN-m (44.2 *0.630) (Force * Lever arm)

SLS combination for crack width (G-I):

Fx = 33.10 kN/m
Fy = 94.55kN (Min)
Mz = 20.86kN-m (33.1 *0.630) (Force * Lever arm)

8.2 Concrete sections studied

The worst case is being studied below with the following assumptions:
The full wheel / rail force is considered as transmitted by one single fastener to plinth.
This fastener is assumed to be the edge one on one plinth segment (depending on the fastener type).
The force spread in plinth is both in transversal and longitudinal directions.

The force spread enables to study the section at the plinth bottom with a maximum flexure moment Mz applied on
it.

Design concrete section after a 45°full spread in longitudinal and transversal directions of rail.

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 Minimum Base plate size = 392mm x 160mm
Effective width in transverse direction = 0.730 m
Effective width in longitudinal direction = 0.550 m
0
(Assuming load dispersion in 45 )
Effective c/s area for load dispersion = 0.730 m x 0.550m

8.3 Reinforced concrete sections check

The section check is carried out in 2 steps:
1. Shear Transfer between plinth bottom and deck.
2. Flexure – Compression on a reinforced concrete section.

8.3.1 First step: Shear Transfer Check

According to clause No. 5.8.4 from AASHTO, we must check the shear transfer between two concrete cast at
different times (deck and plinth).
At the Plinth / Deck interface (full spread section), a minimum steel area is required as follows:
0.35 Acv
Avf (Refer AASHTO cl no. 5.8.4.4, eq. no. 5.8.4.4-1)
fy
which provides the following shear resistance :

Vni cAcv mAvf f y

Where:

c = 0.52 MPa
μ = 0.6 For concrete placed against a
K1 = 0.2 clean concrete surface, free of
K 2= 5.5 MPa laitance, but not intentionally
roughened

Acv = Area of concrete considered to be engaged in interface shear transfer (mm2)

Avf = Area of interface shear reinforcement crossing the shear plane within the area Acv (mm2)
Vui = Factor interface shear force

We check that this minimum area always resists the factored external shear in our study:

Vri Vni Vui (Refer AASHTO cl no. 5.8.4.4, eq. no. 5.8.4.1-1 & 5.8.4.1-2)

The Shear Transfer Check in detailed is carried out here-after.

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8.3.2 Second step: Section under Flexure – Compression Check.
After the force spread, we can study one reinforced section subjected to Flexure and Compression, where the
steel bars crossing the section are modeled.

Remark 1: ULS case

In the ULS case, we remind that the steel area available for the Flexure check must be reduced as certain rebars
work in both Shear and Tension.

So the following rule has to be applied:

- Near compressed part of cross-section:
Steel Area for ULS section check = Steel area crossing the section – Steel area required according to
Shear Transfer Check.
- For tensed part of cross-section:
Steel Area for ULS section check = Steel area crossing the section.

Remark 2: SLS case

For the SLS case, the two following criteria must be fulfilled:
Allowable tensile stress in steel 0.75 * 500 = 375 MPa. (Refer cl no. 5.2.2 of this document)
Allowable compressive stress in concrete 0.5 * 35 = 17.5 MPa. (Refer cl no. 5.2.1 of this document)
Maximum horizontal force in transverse dir. Fx = 44.2 kN/fastener (Refer cl no. 7.1)
Maximum horizontal force in longitudinal dir. Fz = 17.2 kN/fastener (Refer cl no. 7.1)
Hence resultant shear force is 44.2² 17.2² = 47.43 kN
Shear stress due to resultant force is = Force/(c/s area) = 47.43*1000 / (550*730) = 0.118 Mpa
As the shear stress is 0.118 Mpa on the cross-section in the worst case, we can consider that the shear
transfer is done through the concrete/concrete friction at plinth bottom.
So no steel area has to be reduced for Shear Transfer purpose in the Flexure Check of SLS case.

Typical Shear connector Details

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8.4 Shear transfer check (for formula and references refer cl no. 8.3.1)

0 . 35 Acv
Avf
fy

Vni cAcv mAvf f y

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8.5 Section under flexure-compression check
8.5.1 ULS combination:
ULS combination :
Fy = 301.7kN (Max)
Fy = 5.700 kN (Min)
Mz = 48.80 kN-m

Note : For the re-bars in compressed part of cross section, the area remaining for the flexure calculation
after shear transfer is equal to :
6.78/2 – 3.38 = 0.01 cm² (Reinforcement is very small hence ignore compression reinforcement in ULS).

Section 1 Details

Definition
Name Pier
Type Concrete
Material M35
Origin Centre
Dimensions
Depth 550.0mm
Width 730.0mm
Section Area 401500.mm2
Reinforcement Area 339.3mm2
Reinforcement 0.08451%

Section Nodes
Node Y Z
[mm] [mm]
1 365.0 275.0
2 365.0 -275.0
3 -365.0 -275.0
4 -365.0 275.0

Bars
Bar Y Z Diameter Material Type

[mm] [mm] [mm]

1 -255.0 200.0 12.00 FE-500 Steel
2 -255.0 0.0 12.00 FE-500 Steel
3 -255.0 -200.0 12.00 FE-500 Steel

Section Material Properties

Type Concrete
Name M35
Weight Normal Weight
Cube Strength fcu 35.00MPa
Elastic Modulus (short E 20000.MPa
term)
Poisson's Ratio 0.2000

Title : Final Track Plinth Design

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Reinforcement Properties
Name FE-500
fy 500.0MPa
Modulus 200000.MPa

Loading
Applied loads
Load Case N Myy Mzz
[kN] [kNm] [kNm]
1 301.7 0.0 48.80
2 5.700 0.0 48.80

ULS Cases Analysed

Name Loading Pre-stress
Factor
ULS Case 1 L1 1.000
ULS Case 2 L2 1.000

Strength Analysis - Loads

Case N Myy Mzz M
[kN] [kNm] [kNm] [kNm] [°]
1 301.7 0.0 48.80 48.80 -90.00
2 5.700 0.0 48.80 48.80 -90.00

Strength Analysis - Summary

Governing conditions are defined as:
A - reinforcing steel tension strain limit
B - concrete compression strain limit
Effective centroid is reported relative to the reference point.

Case Eff.Eff. N M Mu M/Mu Governing

Centroid Centroid Condition
(y) (z)
[kN] [kNm] [kNm]
Maxima
2 1.434 0.0 5.700 48.80 92.11 0.5298 A: Bar 3
Minima
1 1.394 0.0 301.7 48.80 189.7 0.2573 B: Node 2

Conclusion: As the M/Mu<1, reinforcement provided as shear connector is safe

under flexure.

Title : Final Track Plinth Design

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8.5.2 SLS combination:
SLS combination :
Fy = 190.6kN (Max)
Fy = 5.140 kN (Min)
Mz = 27.85kN-m

SLS combination for crack width (G-I):

Fy = 94.55kN (Min)
Mz = 20.86kN-m
Section 1 Details
Definition
Name Pier
Type Concrete
Material Copy of M35
Origin Centre
Dimensions
Depth 550.0mm
Width 730.0mm
Section Area 401500.mm2
Reinforcement Area 678.6mm2

Section Nodes
Node Y Z
[mm] [mm]
1 365.0 275.0
2 365.0 -275.0
3 -365.0 -275.0
4 -365.0 275.0

Bars
Bar Y Z Diameter Material Type
[mm] [mm] [mm]
1 -255.0 200.0 12.00 FE-500 Steel
2 -255.0 0.0 12.00 FE-500 Steel
3 -255.0 -200.0 12.00 FE-500 Steel
4 235.0 200.0 12.00 FE-500 Steel
5 235.0 0.0 12.00 FE-500 Steel
6 235.0 -200.0 12.00 FE-500 Steel

Section Material Properties

Type Concrete
Name Copy of M35
Weight Normal Weight
Cube Strength fcu 35.00MPa
Elastic Modulus (short E 20000.MPa
term)
Poisson's Ratio 0.2000
Coeff. Thermal Expansion 12.00E-6/°C

Reinforcement Properties
Name FE-500
fy 500.0MPa
Modulus 200000.MPa

Title : Final Track Plinth Design

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Loading
Applied loads
Load Case N Myy Mzz
[kN] [kNm] [kNm]
1 190.6 5.000 27.85
2 5.140 0.0 27.85
3 94.55 5.000 20.86

Serviceability Analysis - Loads

Case N Myy Mzz M M
[kN] [kNm] [kNm] [kNm] [°]
1 190.6 5.000 27.85 28.30 -79.82
2 5.140 0.0 27.85 27.85 -90.00
3 94.55 5.000 20.86 21.45 -76.52

Section Material Stresses/Strains at SLS Loads

Case Bar Coordinates Notes

y z Strain Stress
[mm] [mm] [-] [MPa]
Maxima
2 1 365.0 275.0 107.4E-6 2.148
2 1 365.0 275.0 107.4E-6 2.148
Minima
2 3 -365.0 -275.0 -776.5E-6 0.0
1 3 -365.0 -275.0 -14.35E-6 0.0

Reinforcement Stresses/Strains at SLS Loads

Case Bar Coordinates Notes

y z Strain Stress
[mm] [mm] [-] [MPa]
Maxima
1 4 235.0 200.0 47.36E-6 9.472 FE-500
1 4 235.0 200.0 47.36E-6 9.472 FE-500
Minima
2 3 -255.0 -200.0 -643.3E-6 -128.7 FE-500
2 3 -255.0 -200.0 -643.3E-6 -128.7 FE-500

Crack Widths at SLS Loads

Case Face Point Coordinates Strain Em Strain E1 bt acr Cover Crack width

y zcmin From

[mm] [mm] [mm] [mm] [mm]

3 3 68 -311.0 -96.00 -26.31E-6 -26.31E-6 0.4998 105.1 104.0 Face 3 0.006462

Conclusion:
Compressive stress in concrete is = 2.148Mpa< 0.50Fck Hence ok.
Tensile stress in reinforcement is = 128.6Mpa< 0.75 Fy Hence ok.
Crack width in track plinth is = 0.006462<0.2 (for moderate environment
condition as per IRS:CBC cl no. 10.2.1(a).) Hence OK.

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9 DESIGN OF DERAILMENT GUARD
The derailment force is = 68 kN/m.
Assuming that the derailment force is applied at the top of the derailment guard, then the lever arm for the
derailment force is 0.210m.
The derailment guard is designed as cantilever beam.
The shear force due to derailment force is:
1) ULS = 68*1.75 = 119 kN = 11.9 t.
The Bending moment due to derailment force is:
1) ULS = 68*0.210*1.75 = 25 kN-m = 2.55 t.m.
2) SLS = 68*0.210*1.00 = 15 kN-m =1.50 t.m.

For the derailment force, please refer cl no. 6.1.10 of this

document. And the load factor for the derailment force in
SLS and ULS are as per cl. no. 6.2.
GENERAL DESIGN CRIETERIA

DESIGN FOR SERVICEABILITY LIMIT STATE (SLS)

Grade of concrete M- 35
2
Permissible compressive stress in concrete cbc 0.50fck = 1785 t/m
2
Permissible tensile stress in steel (Fe-500) σst 0.75fy = 38250 t/m
{Refer Table 11 (Stress Limitations for the Serviceability Limit State) of Concrete Bridge Code (IRS)}

Modular ratio = 10
Depth of neutral axis, K=( m* cbc /[m* cbc + st ]) = 0.318
J = 1-(K/3) = 0.894
2
Q = 0.5*K*J* cbc = 253 t/m

DESIGN FOR ULTIMATE LIMIT STATE (ULS)

{Refer cl.15.4.2.2 of Concrete Bridge Code (IRS)}

Mu = ( 0.87 fy ) As Z
2
Mu = 0.15 fck b d
1.1 fy As
Z = [ 1- ]d
fck bd
but not greater than 0.95d

Where
Mu is the ultimate moment of resistance
As is the area of tension reinforcement
b is the width of the section
d is the effective depth to the tension reinforcement
fy is the characteristic strength of the reinforcement
fck is the characteristic strength of the concrete

Title : Final Track Plinth Design

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9.1 Reinforcement calculation for derailment guard

Width of the section 1m

Total depth of the beam 0.18 m

Effective cover to reinforcement 46 mm
ULS SLS

Max. Area of Steel

Required (cm 2)
Effective Depth

Effective Depth

Effective Depth
MOR of C1xAst^2 + C2xAst +
Design B.M.

Area of Steel

Design B.M.

Area of Steel
Member

Total Depth
Governing

Governing
Required

Provided

Required

Required

Required
the Mu =0

(cm 2)

(cm 2)
(t-m)

(t-m)
(m)

(m)

(m)

(m)
section
for
C1 C2
Z=0.95d

Derailment load

Derialment
2.50 0.068 0.18 0.134 2.408 6.97E+05 -5.9E+03 4.44 1.50 0.077 3.274 4.436
Guard

In the above table, the moment of resistance is calculated form actual required reinforcement but the actual
moment of resistance of section with provided reinforcement is as follows:

Actual moment of resistance with provided reinforcement:

2 6
1) From concrete: = 0.15 fck b d = 0.15 x 35 x1000* 134 /9.81 x 10 = 9.6 t-m > Design BM, Hence OK

2) From reinforcement : = 0.87 fy As Z

Z = ( 1-1.1 fy As / (fck b d )) x d
2
Reinforcement provided 12mm dia 100mm c/c, i.e. 1130mm /m
Z = (1-1.1 x 500 x 1130 /(35 x 1000 x 134)) x 134 = 116mm
But Z is not more than = 0.95d = 0.95 x 134 = 127.3 mm

Consider smaller lever arm from above i.e. Z = 116mm

6
Mr = 0.87 fy As Z = 0.87 x 500 x 1130 x 116 / 9.81 x 10 = 5.81 t-m > Design BM, Hence OK

Title : Final Track Plinth Design

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9.2 Shear reinforcement for derailment guard

Width of the section 1m

Total depth of the beam 0.18 m
Clear cover to reinforcement 40 mm
Effective cover to reinforcement 46 mm
Effective depth 134 mm
Grade of concrete 35 Mpa

ULS

%age of tension

stress in conc.

shear stress in
conc. (N/mm 2)
Steel Provided

Ultimate shear
reinforcement
Shear stress

Max. Area of

Depth factor
Design S.F.
Member

Governing

Allowable
(N/mm 2)

(N/mm 2)

(cm 2/m)
Asv/sv
(cm 2)
(t)

Anti derailment R/F not

11.90 0.871 11.3 0.843 0.67 1.39 0.928
guard required

9.3 Reinforcement conclusion

9.3.1 Flexure reinforcement
The flexure reinforcement required in derailment guard is:
2
ULS = 4.44 cm /m,
2
SLS = 3.28 cm /m.
2
The reinforcement provide in derailment guard is 12 dia-100c/c, i.e. 11.30 cm /m.
The flexure reinforcement provided is greater than the reinforcement required, hence OK.

9.3.2 Shear reinforcement

The shear capacity of concrete is greater than actual shear stress. Hence, no shear reinforcement is required.

Title : Final Track Plinth Design

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10 PLINTH REINFORCEMENT DESIGN
10.1 Longitudinal Steel Bars
We choose 10mm Diameter bar with 200 mm spacing from each other, as justified below.

Longitudinal reinforcement concerned by force spread

The maximum length of track plinth is 4.35m and the spacing of fastening system is 0.65m. Hence the no. of
fastening in one unit (4.35/0.65) is 7.
Considering the extreme worst case of loading, the maximum longitudinal force is:
In SLS = 17.2 * 7 = 120.4kN.
In ULS =3.7 * 7 = 25.9kN.

As shown in above figure, the no of longitudinal bars lies in effective width of track plinth is 8Nos. And the forces
in each bar in SLS and ULS are as follows:

Fz ( SLS ) 120.4kN
15.05kN / bar
8 8

Fz (ULS ) 25.9kN
3.24kN / bar
8 8
So each bar is submitted to the following stress in ULS:
3.24kN / bar
s 41.25MPa 500MPa
* 1. 0² / 4
The same check is performed in the SLS case and gives:
15.05kN / bar
s 191.63MPa 375MPa
* 1. 0² / 4

10.2 Transversal frames

We choose 12mm@200mm frames used for confinement reasons.

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10.3 Shrinkage & Temperature check
Minimum reinforcement to control crack spacing shall be the greater of (1) & (2):-
(1) A = f * × A / f
(2) A = f */f × A × ( /2w) × [ R( + ) - 0.5 ]
Where :-
A = area of reinforcement in a given direction to prevent early thermal cracking. This reinforcement should be
distributed evenly around the perimeter of the section.
A = area of effective concrete
f = characteristic tensile strength of reinforcement (N/mm² )
0.7
f * = tensile strength of immature concrete which may be taken as 0.12 (f ) (N/mm² )
f = characteristic cube strength of concrete (N/mm² )
f = average bond strength between reinforcement and immature concrete (N/mm² )
= bar size (nominal diameter) (mm)
w = permissible crack width (mm)
= shrinkage strain, usually not more than 0.5
= ultimate tensile strain capacity of concrete = 200 microstrain
-6
= thermal strain = 0.8 (T + T ), where = 12×10
R = restraint factor
T = short-term fall in temperature from hydration peak to ambient condition
T = long-term fall in temperature from ambient to the seasonal minimum

Minimum reinforcement to control crack spacing based on Equation (1)

h = Beam depth, mm 290 0 0 0 mm

b = Beam w idth, mm 730 0 0 0 mm
cnom = Nominal cover 40 0 0 0 mm (Nominal concrete cover to r/f)
Ldia = Dia. of link 0 0 0 0 mm
c ten = Cover to reinf. 40 0 0 0 mm (Cover to tension reinforcement)
A = Eff. concrete area 211700 0 0 0 mm² w here
Ac = h ´ b - [ max( 0 , (h - 500)) max( 0 , (b - 500 ))]
f = (N/mm²) 35 N/mm²
0.7
f * = 0.12 (f ) 1.45 N/mm²
f = (N/mm²) 500 N/mm²
A = f *×A /f 612.0 0.0 0.0 0.0 mm²

Title : Final Track Plinth Design

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 Reinforcement for early thermal cracking based on Equation (2) as per BD 28/87

A = Eff. concrete area 211700 0 0 0 mm²

= Dia. of rebar 10 mm
w = Crack width 0.1 mm
= 0.5 0.0001
= ult. tensile strain 0.0002
T = 43 oC
T = 20 oC Normally taken as 20 oC for summer concreting
= thermal strain 0.000605 Thermal strain = 0.8 (T1 + T2), w here = 12x10-6

R = restraint factor 0.2 See Table 2 of BD 28/87

f */f = 0.67 For Type 2 deformed Bars

Restraint Condition Restraint

Factor
( Table 2 of BD 28/87 )
Area of steel required, A = f */f × A × ( /2w) × [ R( + ) - 0.5 ] External : R

A = (mm²) 290.5 0.0 0.0 0.0 mm² 0.2

A = (mm²) 612.0 0.0 0.0 0.0 mm² 0.5

As_therm_reqd =Max (As1 , As2) 612.0 0.0 0.0 0.0 mm² 0.6
0.8
1.0

Perimeter for early thermal cracking steel; Pt = 2 (h - 2 cten) + 2 (b - 2 cten ) Internal : 0.5

Pt = (mm) 1720.0 mm
A s_per_m = As_therm_reqd /Pt 355.8 mm²/m ==> Area of steel required per metre
= Dia. (mm) Ø10 mm
Spacing =( 2 /4) / As_per_m 220.7 mm ==> Max. spacing of bars
Spacing =
(mm) 200 mm
A = (mm²) 392.7 mm²
O.K. !

2
Reinforcement required due to shrinkage and temperature for in longitudinal direction is 612 mm and the
2
reinforcement provided in longitudinal direction is 10 dia. bar 10nos. i.e. 785mm , Hence OK.

Title : Final Track Plinth Design

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11 TRACK PLINTH WITH CHECK RAIL

The check rail is only provided in canted (super elevated) track to limit the lateral wheel displacement & avoid
derailment which may occur due to the centrifugal force acting on it, so effectively check rail is subjected to only
derailment force for which track plinth need to be checked:
As per cl no. 6.1.10, the intensity of derailment force is 68 KN/m, and the spacing of fastener for check rail is
0.65m. Hence the effective horizontal force in transverse direction acting over check rail per fastener is,
FX = 68 x 1.3 = 88.4 KN/Fastener (1.3m fastener spacing at end of track plinth)

It is assumed that the horizontal force is acting at rail top. Total height of check rail top from the bottom of track
plinth is (321+199) 520mm as shown in following figure,

Typical Cross Section for Track Plinth with Check Rail

Hence lever arm for horizontal forces = (321+199) = 520 mm ≈ 0.550 m

At the time of derailment, it is assumed that there is no vertical force acting over track plinth and only check rail is
subjected to derailment force.

Force summery is as follows:

SLS ULS
FY 0 kN 0 kN
FX 88.4 kN 154.7 kN (88.4*1.75)
MX 48.7 kN-m 85.10 kN-m (154.7*0.550)
Fz 17.2 kN 3.700 kN (Refer cl no. 7.1)

For check rail track plinth, it is assumed that LWR & braking traction force for check rail is same as for main rail
but in reality there is no longitudinal force acting over check rail. It is considered on conservative side.

Title : Final Track Plinth Design

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 Load dispersion for check rail in both directions

11.1 Shear Check SLS case

For the SLS case, the two following criteria must be fulfilled:
Allowable tensile stress in steel 0.75 * 500 = 375 MPa. (Refer cl no. 5.2.2 of this document)
Allowable compressive stress in concrete 0.5 * 35 = 17.5 MPa. (Refer cl no. 5.2.1 of this document)
Maximum horizontal force in transverse dir. Fx = 88.4 kN/fastener (Refer cl no. 11)
Maximum horizontal force in longitudinal dir. Fz = 17.2 kN/fastener (Refer cl no. 11)

Hence resultant shear force is 88 .5² 17.2² = 90.06 kN

Shear stress due to resultant force is = Force/(c/s area) = 90.06*1000 / (706*645) = 0.198Mpa
As the shear stress is 0.198Mpa on the cross-section in the worst case, we can consider that the shear
transfer is done through the concrete/concrete friction at plinth bottom.

Title : Final Track Plinth Design

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11.2 Shear Check ULS case

0 . 35 Acv
Avf
fy

Vni cAcv mAvf f y

Title : Final Track Plinth Design

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11.3 Section under flexure-compression check
11.3.1 ULS combination:
Fy = 0 kN
Mz = 85.1kN.m

Note : For the re-bars in compressed part of cross section, the area remaining for the flexure calculation
after shear transfer is equal to :
0.785*4 –3.84 = -0.70 cm² (Ignore compression reinforcement and reduced Tension reinforcement by
2
0.70cm i.e. ignore 10dia. 2 no. of bar in tension for ULS case).

Section 1 Details
Definition
Name Pier
Type Concrete
Material Copy of M35
Origin Centre
Dimensions
Depth 706.0mm
Width 645.0mm
Section Area 455400.mm2
Reinforcement Area 609.5mm2
Reinforcement 0.1338%
Section Nodes
Node Y Z
[mm] [mm]
1 322.5 353.0
2 322.5 -353.0
3 -322.5 -353.0
4 -322.5 353.0

Bars
Bar Y Z Diameter Material Type

[mm] [mm] [mm]

1 -127.5 300.0 12.00 FE-500 Steel
2 -127.5 100.0 12.00 FE-500 Steel
3 -127.5 -100.0 12.00 FE-500 Steel
4 -127.5 -300.0 12.00 FE-500 Steel
5 -67.50 100.0 10.00 FE-500 Steel
6 -67.50 -100.0 10.00 FE-500 Steel

Section Material Properties

Type Concrete
Name Copy of M35
Elastic Modulus (short E 20000.MPa
term)
Poisson's Ratio 0.2000

Reinforcement Properties
Name FE-500
fy 500.0MPa
Modulus 200000.MPa

Title : Final Track Plinth Design

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Loading
Applied loads
Load Case N Myy Mzz
[kN] [kNm] [kNm]
1 0.0 0.0 85.10

Strength Analysis - Loads

Case N Myy Mzz M
[kN] [kNm] [kNm] [kNm] [°]
1 0.0 0.0 85.10 85.10 -90.00

Strength Analysis - Summary

Governing conditions are defined as:
A - reinforcing steel tension strain limit
B - concrete compression strain limit
Effective centroid is reported relative to the reference point.

Case Eff.Eff. N M Mu M/Mu Governing

Centroid Centroid Condition
(y) (z)
[kN] [kNm] [kNm]
Maxima
1 - - 0.0 85.10 112.2 0.7587 A: Bar 1

Conclusion: As the M/Mu<1, reinforcement provided as shear connector is safe

under flexure.

Title : Final Track Plinth Design

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11.3.2 SLS combination:
Fy = 0 kN
Mz = 48.70kN-m

Section 1 Details
Definition

Name Pier
Type Concrete
Material Copy of M35
Origin Centre
Dimensions
Depth 706.0mm
Width 645.0mm
Section Area 455400.mm2
Reinforcement Area 1081.mm2
Reinforcement 0.2373%

Section Nodes
Node Y Z
[mm] [mm]
1 322.5 353.0
2 322.5 -353.0
3 -322.5 -353.0
4 -322.5 353.0

Bars
Bar Y Z Diameter Material Type

[mm] [mm] [mm]

1 -127.5 300.0 12.00 FE-500 Steel
2 -127.5 100.0 12.00 FE-500 Steel
3 -127.5 -100.0 12.00 FE-500 Steel
4 -127.5 -300.0 12.00 FE-500 Steel
5 -67.50 300.0 10.00 FE-500 Steel
6 -67.50 100.0 10.00 FE-500 Steel
7 -67.50 -100.0 10.00 FE-500 Steel
8 -67.50 -300.0 10.00 FE-500 Steel
9 232.5 300.0 10.00 FE-500 Steel
10 232.5 100.0 10.00 FE-500 Steel
11 232.5 -100.0 10.00 FE-500 Steel
12 232.5 -300.0 10.00 FE-500 Steel

Section Material Properties

Type Concrete
Name Copy of M35
Cube Strength fcu 35.00MPa
Elastic Modulus (short E 20000.MPa
term)
Poisson's Ratio 0.2000

Title : Final Track Plinth Design

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Reinforcement Properties
Name FE-500
fy 500.0MPa
Modulus 200000.MPa

Loading
Applied loads
Load Case N Myy Mzz
[kN] [kNm] [kNm]
1 0.0 0.0 48.70

Serivceability Analysis - Loads

Case N Myy Mzz M
[kN] [kNm] [kNm] [kNm] [°]
1 0.0 0.0 48.70 48.70 -90.00

Section Material Stresses/Strains at SLS Loads

Case Bar Coordinates
y z Strain Stress
[mm] [mm] [-] [MPa]
Maxima
1 1 322.5 353.0 201.6E-6 4.032
1 1 322.5 353.0 201.6E-6 4.032
Minima
1 3 -322.5 -353.0 -0.001309 0.0
1 3 -322.5 -353.0 -0.001309 0.0
Reinforcement Stresses/Strains at SLS Loads
Case Bar Coordinates
y z Strain Stress
[mm] [mm] [-] [MPa]
Maxima
1 9 232.5 300.0 -9.252E-6 -1.850 FE-500
1 9 232.5 300.0 -9.252E-6 -1.850 FE-500
Minima
1 3 -127.5 -100.0 -852.6E-6 -170.5 FE-500
1 3 -127.5 -100.0 -852.6E-6 -170.5 FE-500

Conclusion:
Compressive stress in concrete is =4.032MPa< 0.5 Fck Hence ok.
Tensile stress in reinforcement is = 170.5MPa< 0.75 Fy Hence ok.

Typical Shear connector details

Title : Final Track Plinth Design
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11.4 Reinforcement calculation for derailment guard

The derailment guard for track plinth with check rail is same as derailment guard for standard track plinth, Hence
for the design of derailment guard, please refer cl .no. 9.1. The reinforcement arrangement for derailment guard
for track plinth with check rail is as show in following fig.

Typical reinforcement arrangement for derailment guard with check rail

Title : Final Track Plinth Design

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12 PLINTH REINFORCEMENT DESIGN
12.1 Longitudinal Steel Bars
We choose 10mm Diameter bar with 200 mm spacing from each other, as justified below.

- 12 NOS

Longitudinal reinforcement

The maximum length of track plinth is 4.35m and the spacing of fastening system is 0.65m. Hence the no. of
fastening in one unit (4.35/0.65) is 7.
Considering the extreme worst case of loading, the maximum longitudinal force is:
In SLS, = 17.2 * 7 = 120.4 KN.
In ULS, = 3.7 * 7 = 25.9 KN.

As shown in above figure, the no of longitudinal bars lies in effective width of track plinth is 5 Nos. And the forces
in each bar in SLS and ULS are as follows:

Fz ( SLS ) 120.4kN
24.08kN / bar
5 5

Fz (ULS ) 25.9kN
5.18kN / bar
5 5
So each bar is submitted to the following stress in ULS:
5.18kN / bar
s 65.95MPa 500 MPa
* 1.0² / 4
The same check is performed in the SLS case and gives:

24.08kN / bar
s 306.6MPa 375MPa
* 1. 0² / 4

12.2 Transversal frames

We choose 12mm@200mm frames used for confinement reasons.

Title : Final Track Plinth Design

Ref.: HM-LE-PRWIDE-M0601-DOC-DC-TRW-00-101-C Page : 37/ 37
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